In the circuit shown, the galvanometer G of resistance 60 Ω is shunted by a resistance r = 0.02 Ω . The current through R is nearly 1 A . The value of resistance R is (in ohm) is nearly

Question

In the circuit shown, the galvanometer G of resistance 60 Ω is shunted by a resistance r = 0.02 Ω . The current through R is nearly 1 A . The value of resistance R is (in ohm) is nearly <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/bf635fa46eb002ab682e30bcde068e4b-.png" /> (A)  1.00 Ω  (B)  5.00 Ω  (C)  11.00 Ω  (D)  60.00 Ω

Tardigrade Admin · Accepted Answer

Equivalent resistance of the combination = (GS/G+S) = (60 × 0.02/60+ 0.02) = 0.0199 Total resistance = R + 0.0199 V = iR' 5 = 1 × R + 0.0199 R = 5- 0.0199 = 4.98 ≈ 5.00 Ω

Q. In the circuit shown, the galvanometer $G$ of resistance $60 Ω$ is shunted by a resistance $r = 0.02 Ω$ . The current through $R$ is nearly $1 A$ . The value of resistance $R$ is (in ohm) is nearly

$1.00 Ω$

$5.00 Ω$

$11.00 Ω$

$60.00 Ω$

Equivalent resistance of the combination
$= \frac{GS}{G + S} = \frac{60 \times 0.02}{60 + 0.02}$
$= 0.0199$
Total resistance $= R + 0.0199$
$V = i R^{'}$
$5 = 1 \times R + 0.0199$
$R = 5 - 0.0199 = 4.98 \approx 5.00 Ω$

Q. In the circuit shown, the galvanometer G of resistance 60Ω is shunted by a resistance r=0.02Ω . The current through R is nearly 1A . The value of resistance R is (in ohm) is nearly

1.00Ω

5.00Ω

11.00Ω

60.00Ω